数学分析_复旦_欧阳光中陈传璋第三版3版上下册课后习题答案解析(下)
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x→0
=
lim
x→0
ln x 1
=
lim
x→0
x −1
= − lim x = 0§u´ lim xsin x = 1
x→0
x→0
x
x2
(18)
1
lim x 1−x
= e § lim x→1
ln x 1−x
x→1
1
lim
x→1
ln x 1−x
= − lim
x→1
1 x
=
−1§u´ lim x 1 − x
4 cos2 x
lim
x→∞
[2 + cos x(1 + 2x + sin x)]esin x
lim
+
=
x→0
24
24
1
6
ax − bx
ax ln a − bx ln b
a
(9) lim
= lim
= ln a − ln b = ln (a = 0, b = 0)
x→0 x
x→0
1
b
x−1
1
(10) lim
x→1
ln x
= lim
x→1
1
=1
x
(11) lim ax − xa = lim ax ln a − axa−1 = aa(ln a − 1)
=
x→0
12x2
sin(sin x) cos3 x + 3 cos(sin x) sin 2x + sin(sin x) cos x − sin x
lim
2
=
x→0
24x
cos(sin x) cos4 x − 3 sin(sin x) sin 2x cos x + 3 cos(sin x) cos 2x cos(sin x) cos2 x − sin(sin x) sin x − cos x
lnc x =
lim
yc = 0(d(4)
)
x→+∞ xb
y→+∞ eby
1
(1 + x) x − e
1
(15) lim
= lim (1+x) x
x→0
x
x→0
−e
2
1
1
x − (1 + x) ln(1 + x)
1 − 1 − ln(1 + x)
− ln(1 + x) = e lim
= e lim
=
1+
=
lim
x→∞
1
−
x cos x
=1
x
(3) éuØÓ
S
µxn
=
2nπ
+
π 2
9xn
=
2nπ(n
=
1, 2, · · ·
)§
4•Ø•3.
^â7ˆ{K¦)§k lim 2x + sin 2x = x→∞ (2x + sin x)esin x
2 + 2 cos 2x
lim
=
x→∞ (2 + cos x + 2x cos x + sin x cos x)esin x
x(1 + x) x2
x→0
x2
x→0
2x
(16) -y = ln x,Kx = ey§u´ lim xb lnc x = lim ebyyc = lim yc = 0(d(4) )
x→0
y→−∞
y→−∞ e−by
(17)
lim
xsin x
=
lim
ex→0
sin
x
ln
x
§
x→0
1
lim sin x ln x
xb =0
x→∞ eax
†!müà
x → ∞ž§
(5) lim
x→1
1
1
1
−
ln x x − 1
1
x − 1 − ln x = lim
x→1 (x − 1) ln x
1−
=
lim
x→1
ln x
+
x x−
1
x−1 = lim
x→1 x ln x + x − 1
1 = lim
x→1 ln x + 1 + 1
x
x→+0
x y→+∞ y
y→+∞ y ln y
lim
x→+0
ln 1 x
x
=1
2. Á`²e ¼êØU^â7ˆ{K¦4•µ
x2 sin 1
(1) lim
x
x→0 sin x
x + sin x (2) lim
x→∞ x − cos x
2x + sin 2x (3) lim
x→∞ (2x + sin x)esin x
=
x
2
x
π−x
−1
(6)
lim (π − x) tan
x→π
2
=
lim
x→π
cot
x
2
=
lim
x→π
1 −
csc2
x
22
=2
ln(cos ax)
−a tan ax a tan ax a a sec2 ax a2
(7)
lim x→0 ln(cos bx)
= lim
x→0
−b tan bx
=
b
lim
x→0
tan bx
=
b
lim
x→0
b sec2 bx
=
b2 (b = 0)
cos(sin x) − cos x
− sin(sin x) cos x + sin x
(8) lim
x→0
x4
= lim
x→0
4x3
− cos(sin x) cos2 x + sin(sin x) sin x + cos x
= lim
x→a x − a
x→a
1
√
1 − 2 sin x
−2 cos x
3
(12) lim
= lim
=
x→
π 6
cos 3x
x→
π 6
−3 sin 3x
3
1
ln x
(13) lim
= lim
x
sin2 x
= − lim
=0
x→0 cot x x→0 − csc2 x
x→0 x
(14) -y = ln x,Kx = ey§u´ lim
101
(4) b•
ê§ lim
x→∞
xb eax
=
lim
x→∞
bxb−1 aeax
Biblioteka Baidu
=
··· =
lim
x→∞
b! abeax
=0
bؕ
ê§K[b]
b
<
[b]+1§u´
|x|[b] eax
|x|b eax
<
|x|[b]+1 eax (|x|
> 1)§
þ¡®y²§‚ 4••0§Ïd§¥m 4•••0.
l
§é?¿a, b§þk lim
(x2 − 1) sin x
(4) lim x→1 ln
1 + sin π x
2
)µ
x2 sin 1
1
1
2x sin − cos
1 cos
(1) Ï
x ©f!©1Óžéx¦ ê§
x
x§
x x → 0ž4•Ø•3§Ïdâ
sin x
cos x
cos x
7ˆ{KØU·^§
´
4•´•3
x2 sin 1
"¯¢þ§k lim
x = lim
x
· x sin 1 = 0
x→0 sin x
x→0 sin x
x
(2) Ï x + sin x ©f!©1Óžéx¦ ê§ 1 + cos x § x → ∞žd¼ê4•Ø•3§Ïdâ7ˆ{
x − cos x
1 + sin x
sin x
KØU·^§
´
4•´•3
"¯¢þ§k lim x + sin x x→∞ x − cos x
x→1
=
1 e
1
1
ex − x − 1
ex − 1
ex
1
(19) lim
x→0
x − ex − 1
= lim
x→0
x(ex − 1)
=
lim
x→0
=
ex
−
1+
xex
=
lim
x→0
2ex
+
xex
= 2
102
(20) lim
x→+0
1 ln
x
1
x
=
lim x lnln x→+0
e
x
-y = 1 §K lim x ln(ln 1 ) = lim ln(ln y) = lim 1 = 0§l
=
lim
x→0
ln x 1
=
lim
x→0
x −1
= − lim x = 0§u´ lim xsin x = 1
x→0
x→0
x
x2
(18)
1
lim x 1−x
= e § lim x→1
ln x 1−x
x→1
1
lim
x→1
ln x 1−x
= − lim
x→1
1 x
=
−1§u´ lim x 1 − x
4 cos2 x
lim
x→∞
[2 + cos x(1 + 2x + sin x)]esin x
lim
+
=
x→0
24
24
1
6
ax − bx
ax ln a − bx ln b
a
(9) lim
= lim
= ln a − ln b = ln (a = 0, b = 0)
x→0 x
x→0
1
b
x−1
1
(10) lim
x→1
ln x
= lim
x→1
1
=1
x
(11) lim ax − xa = lim ax ln a − axa−1 = aa(ln a − 1)
=
x→0
12x2
sin(sin x) cos3 x + 3 cos(sin x) sin 2x + sin(sin x) cos x − sin x
lim
2
=
x→0
24x
cos(sin x) cos4 x − 3 sin(sin x) sin 2x cos x + 3 cos(sin x) cos 2x cos(sin x) cos2 x − sin(sin x) sin x − cos x
lnc x =
lim
yc = 0(d(4)
)
x→+∞ xb
y→+∞ eby
1
(1 + x) x − e
1
(15) lim
= lim (1+x) x
x→0
x
x→0
−e
2
1
1
x − (1 + x) ln(1 + x)
1 − 1 − ln(1 + x)
− ln(1 + x) = e lim
= e lim
=
1+
=
lim
x→∞
1
−
x cos x
=1
x
(3) éuØÓ
S
µxn
=
2nπ
+
π 2
9xn
=
2nπ(n
=
1, 2, · · ·
)§
4•Ø•3.
^â7ˆ{K¦)§k lim 2x + sin 2x = x→∞ (2x + sin x)esin x
2 + 2 cos 2x
lim
=
x→∞ (2 + cos x + 2x cos x + sin x cos x)esin x
x(1 + x) x2
x→0
x2
x→0
2x
(16) -y = ln x,Kx = ey§u´ lim xb lnc x = lim ebyyc = lim yc = 0(d(4) )
x→0
y→−∞
y→−∞ e−by
(17)
lim
xsin x
=
lim
ex→0
sin
x
ln
x
§
x→0
1
lim sin x ln x
xb =0
x→∞ eax
†!müà
x → ∞ž§
(5) lim
x→1
1
1
1
−
ln x x − 1
1
x − 1 − ln x = lim
x→1 (x − 1) ln x
1−
=
lim
x→1
ln x
+
x x−
1
x−1 = lim
x→1 x ln x + x − 1
1 = lim
x→1 ln x + 1 + 1
x
x→+0
x y→+∞ y
y→+∞ y ln y
lim
x→+0
ln 1 x
x
=1
2. Á`²e ¼êØU^â7ˆ{K¦4•µ
x2 sin 1
(1) lim
x
x→0 sin x
x + sin x (2) lim
x→∞ x − cos x
2x + sin 2x (3) lim
x→∞ (2x + sin x)esin x
=
x
2
x
π−x
−1
(6)
lim (π − x) tan
x→π
2
=
lim
x→π
cot
x
2
=
lim
x→π
1 −
csc2
x
22
=2
ln(cos ax)
−a tan ax a tan ax a a sec2 ax a2
(7)
lim x→0 ln(cos bx)
= lim
x→0
−b tan bx
=
b
lim
x→0
tan bx
=
b
lim
x→0
b sec2 bx
=
b2 (b = 0)
cos(sin x) − cos x
− sin(sin x) cos x + sin x
(8) lim
x→0
x4
= lim
x→0
4x3
− cos(sin x) cos2 x + sin(sin x) sin x + cos x
= lim
x→a x − a
x→a
1
√
1 − 2 sin x
−2 cos x
3
(12) lim
= lim
=
x→
π 6
cos 3x
x→
π 6
−3 sin 3x
3
1
ln x
(13) lim
= lim
x
sin2 x
= − lim
=0
x→0 cot x x→0 − csc2 x
x→0 x
(14) -y = ln x,Kx = ey§u´ lim
101
(4) b•
ê§ lim
x→∞
xb eax
=
lim
x→∞
bxb−1 aeax
Biblioteka Baidu
=
··· =
lim
x→∞
b! abeax
=0
bؕ
ê§K[b]
b
<
[b]+1§u´
|x|[b] eax
|x|b eax
<
|x|[b]+1 eax (|x|
> 1)§
þ¡®y²§‚ 4••0§Ïd§¥m 4•••0.
l
§é?¿a, b§þk lim
(x2 − 1) sin x
(4) lim x→1 ln
1 + sin π x
2
)µ
x2 sin 1
1
1
2x sin − cos
1 cos
(1) Ï
x ©f!©1Óžéx¦ ê§
x
x§
x x → 0ž4•Ø•3§Ïdâ
sin x
cos x
cos x
7ˆ{KØU·^§
´
4•´•3
x2 sin 1
"¯¢þ§k lim
x = lim
x
· x sin 1 = 0
x→0 sin x
x→0 sin x
x
(2) Ï x + sin x ©f!©1Óžéx¦ ê§ 1 + cos x § x → ∞žd¼ê4•Ø•3§Ïdâ7ˆ{
x − cos x
1 + sin x
sin x
KØU·^§
´
4•´•3
"¯¢þ§k lim x + sin x x→∞ x − cos x
x→1
=
1 e
1
1
ex − x − 1
ex − 1
ex
1
(19) lim
x→0
x − ex − 1
= lim
x→0
x(ex − 1)
=
lim
x→0
=
ex
−
1+
xex
=
lim
x→0
2ex
+
xex
= 2
102
(20) lim
x→+0
1 ln
x
1
x
=
lim x lnln x→+0
e
x
-y = 1 §K lim x ln(ln 1 ) = lim ln(ln y) = lim 1 = 0§l